Question

\(\int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx =\)

• A. \(2 \cos x + 2x \cos \alpha + c\), where c is the constant of integration.
• B. \(2 \cos x - 2x \cos \alpha + c\), where c is the constant of integration.
• C. \(2 \sin x + 2x \cos \alpha + c\), where c is the constant of integration.
• D. \(2 \sin x + 2x \sin \alpha + c\), where c is the constant of integration.

Hint:

When an integrand contains \(\cos 2x - \cos 2\alpha\), try factorizing it first. It often cancels the denominator immediately.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The given integral involves trigonometric functions with different multiples of the variable \(x\) and constant \(\alpha\).
To simplify the fraction, we need to convert the double-angle terms in the numerator to single-angle terms.
Step 2: Key Formula or Approach:
We use the double-angle formula for cosine:
\[ \cos 2\theta = 2\cos^2 \theta - 1 \] Additionally, we use the algebraic identity for the difference of squares:
\[ a^2 - b^2 = (a-b)(a+b) \] Step 3: Detailed Explanation:
Substitute the double-angle formula into the numerator:
\[ \int \frac{(2\cos^2 x - 1) - (2\cos^2 \alpha - 1)}{\cos x - \cos \alpha} dx \] \[ = \int \frac{2\cos^2 x - 1 - 2\cos^2 \alpha + 1}{\cos x - \cos \alpha} dx \] \[ = \int \frac{2(\cos^2 x - \cos^2 \alpha)}{\cos x - \cos \alpha} dx \] Factor the numerator using the difference of squares:
\[ = 2 \int \frac{(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} dx \] Cancel the common term in the numerator and denominator:
\[ = 2 \int (\cos x + \cos \alpha) dx \] Now, integrate term by term. Note that \(\cos \alpha\) is a constant with respect to \(x\):
\[ = 2 [ \sin x + x \cos \alpha ] + c \] \[ = 2 \sin x + 2x \cos \alpha + c \] Step 4: Final Answer:
The value of the integral is \(2 \sin x + 2x \cos \alpha + c\).